{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4 : First Law of Thermodynamics" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.1 page no : 119" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Calculate the change in internal energy of the working fluid stating whether it is a gain or loss.\n", "\n", "# Variables\n", "Q = -50.; \t\t\t#kJ/kg\n", "W = -100.; \t\t\t#kJ/kg\n", "\n", "# Calculations\n", "dU = Q-W;\n", "\n", "# Results\n", "print (\"gain in internal energy = \"),(dU),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "gain in internal energy = 50.0 kJ/kg\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.2 page no : 119" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "calculate the heat flow to and from the cylinder\n", "'''\n", "\n", "# Variables\n", "u1 = 450.; \t\t\t#kJ/kg\n", "u2 = 220; \t\t\t#kJ/kg\n", "W = 120; \t\t\t#kJ/kg\n", "\n", "# Calculations\n", "Q = (u2-u1) + W;\n", "\n", "# Results\n", "print (\"Heat rejected by air = \"),(-Q),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat rejected by air = 110.0 kJ/kg\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.3 page no : 119" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the heat transferred from the nitrogen to the surroundings.\n", "'''\n", "\n", "# Variables\n", "m = 0.3; \t\t\t#kg\n", "cv = 0.75; \t\t\t#kJ/kg.K\n", "T1 = 313.; \t\t\t#K\n", "T2 = 433.; \t\t\t#K\n", "W = -30.; \t\t\t#kJ\n", "\n", "# Calculations\n", "dU = m*cv*(T2-T1);\n", "Q = dU + W;\n", "\n", "# Results\n", "print (\"Heat rejected during the process = \"),(-Q),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat rejected during the process = 3.0 kJ\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.4 page no : 120" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "How much did the internal energy of the gas change ?\n", "'''\n", "# Variables\n", "p1 = 105.; \t\t\t#kPa\n", "V1 = 0.4; \t\t\t#m**3\n", "p2 = p1;\n", "V2 = 0.20; \t\t\t#m**3\n", "Q = -42.5; \t\t\t#kJ\n", "\n", "# Calculations\n", "W = p1*(V2-V1);\n", "dU = Q-W;\n", "\n", "# Results\n", "print (\"change in internal energy = \"),(dU),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "change in internal energy = -21.5 kJ\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.6 Page no :121" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine : (i) Work done ;\n", "(ii) Change in internal energy ; and\n", "(iii) Heat transferred\n", "'''\n", "import math\n", "\n", "# Variables\n", "p1 = 10.**5 # Pa\n", "T1 = 25 + 273 # K\n", "p2 = 5 * 10**5 # Pa\n", "T2 = T1 \n", "\n", "# Calculations and Result\n", "print \"(i) For isothermal process :\"\n", "W12 = -p1 * (1.8)* math.log(p1/p2)\n", "print \"Work done on the air = %.3e kJ/kg.\"%W12\n", "\n", "print \"(ii) Since temperature is constant,\"\n", "print \"u2 \u2013 u1 = 0\"\n", "print \"Change in internal energy = zero.\"\n", "\n", "print \"(iii) Again,\"\n", "Q12 = 0 + W12\n", "print \"Heat rejected = %.3e kJ/kg.\"%Q12\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) For isothermal process :\n", "Work done on the air = 2.897e+05 kJ/kg.\n", "(ii) Since temperature is constant,\n", "u2 \u2013 u1 = 0\n", "Change in internal energy = zero.\n", "(iii) Again,\n", "Heat rejected = 2.897e+05 kJ/kg.\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.7 page no : 122" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the quantity of heat added to the system.\n", "'''\n", "\n", "\n", "# Variables\n", "W_12 = -82.; \t\t\t#kJ\n", "Q_12 = -45.; \t\t\t#kJ\n", "dU_12 = Q_12 - W_12;\n", "W_21 = 100.; \t\t\t#kJ\n", "dU_21 = -dU_12;\n", "\n", "# Calculations\n", "Q_21 = dU_21 + W_21;\n", "\n", "# Results\n", "print (\"Heat added to the system = \"),(Q_21),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat added to the system = 63.0 kJ\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.8 page no : 123" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine : \n", "(i) Work done ;\n", "(ii) Change in internal energy of the system.\n", "'''\n", "\n", "# Variables\n", "Q2 = 9000.; \t\t\t#kJ\n", "Q1 = 3000.; \t\t\t#kJ\n", "Q = Q1-Q2; \n", "W = 0;\n", "\n", "# Calculations\n", "dU = W-Q;\n", "\n", "# Results\n", "print (\"Work done = \"),(W)\n", "\n", "print (\"Change in internal energy = \"),(dU),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done = 0\n", "Change in internal energy = 6000.0 kJ\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.9 page no : 124" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine \u2206U, \u2206PE, \u2206KE, Q and W, when\n", "(i) the stone is about to enter the water,\n", "(ii) the stone has come to rest in the tank, and\n", "(iii) the heat is transferred to the surroundings in such an amount that the stone and water\n", "come to their initial temperature.\n", "'''\n", "\n", "# Variables\n", "m = 20.; \t\t\t#kg\n", "g = 9.81; \t\t\t#m/s**2\n", "z2 = 0.;\n", "z1 = 15.;\n", "\n", "# Calculations and Results\n", "print (\"(i) When the stone is about to enter the water\")\n", "Q = 0\n", "W = 0\n", "dU = 0\n", "PE = m*g*(z2-z1)\n", "KE = -PE\n", "print \"\u2206 PE = %.3f\"%KE,\"J\"\n", "\n", "print (\"(ii) When the stone dips into the math.tank and comes to rest\")\n", "Q = 0\n", "W = 0\n", "KE = 0\n", "PE = m*g*(z2-z1)\n", "dU = -PE\n", "print \"\u2206U = %.3f\"%dU\n", "\n", "print (\"(iii) When the water and stone come to their initial temperature\")\n", "W = 0\n", "KE = 0\n", "Q = -dU\n", "print \"Q = %.3f\"%Q\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) When the stone is about to enter the water\n", "\u2206 PE = 2943.000 J\n", "(ii) When the stone dips into the math.tank and comes to rest\n", "\u2206U = 2943.000\n", "(iii) When the water and stone come to their initial temperature\n", "Q = -2943.000\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.10 page no : 125" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) How much will be the heat that flows into the system along path lnm if the work done\n", "is 21 kJ ?\n", "(ii) When the system is returned from m to l along the curved path, the work done on the\n", "system is 42 kJ. Does the system absorb or liberate heat, and how much of the heat is absorbed\n", "or liberated ?\n", "(iii) If U l = 0 and U n = 84 kJ, find the heat absorbed in the processes ln and nm.\n", "'''\n", "\n", "# Variables\n", "Q_lqm = 168.; \t\t\t#kJ\n", "W_lqm = 64.; \t\t\t#kJ\n", "dU_lm = Q_lqm - W_lqm;\n", "W_lnm = 21.; \t\t\t#kJ\n", "W_ml = -42.; \t\t\t#kJ\n", "\n", "# Calculations and Results\n", "Q_lnm = dU_lm + W_lnm;\n", "print (\"(i)Q_lnm = \"),(Q_lnm), (\"kJ\")\n", "\n", "Q_ml = W_ml - dU_lm;\n", "print (\"(ii)Q_ml = \"),(Q_ml),(\"kJ\")\n", "\n", "W_ln = 21.; \t\t\t#kJ\n", "dU_ln = 84.; \t\t\t#kJ\n", "Q_ln = dU_ln + W_ln;\n", "Q_nm = Q_lnm-Q_ln;\n", "print (\"(iii)Q_nm = \"), (Q_nm), (\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)Q_lnm = 125.0 kJ\n", "(ii)Q_ml = -146.0 kJ\n", "(iii)Q_nm = 20.0 kJ\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.11 page no : 126" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "What will be the change of internal energy of the system when its temperature changes\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "T1 = 55.; \t\t\t#0C\n", "T2 = 95.; \t\t\t#0C\n", "\n", "# Calculations\n", "def f1( T): \n", "\t return 200\n", "W = quad(f1, T1, T2)[0]\n", "\n", "\n", "def f2( T): \n", "\t return 160\n", "Q = quad(f2, T1, T2)[0]\n", "dU = Q-W;\n", "\n", "# Results\n", "print (\"change in internal energy = \"),(dU/10**3),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "change in internal energy = -1.6 kJ\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.12 page no : 127" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Completeing the table\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "Q = -340.; \t\t\t#kJ\n", "n = 200.; \t\t\t#cycles/min\n", "\n", "#For Process 1-2\n", "W_12 = 4340.; \t\t\t#kJ/min\n", "Q_12 = 0.;\n", "\n", "# Calculations and Results\n", "dE_12 = Q_12-W_12;\n", "print (\"dE_12 = \"),(dE_12),(\"kJ/min\")\n", "\n", "#For process 2-3\n", "Q_23 = 42000.; \t\t\t#kJ/min\n", "W_23 = 0;\n", "\n", "dE_23 = Q_23-W_23;\n", "print (\"dE_23 = \"),(dE_23),(\"kJ/min\")\n", "\n", "#For process 3-4\n", "Q_34 = -4200.; \t\t\t#kJ/min\n", "dE_34 = -73200.; \t\t\t#kJ/min\n", "\n", "W_34 = Q_34-dE_34;\n", "print (\"W_34 = \"), (W_34), (\"kJ/min\")\n", "\n", "#For process 4-1\n", "Q_41 = Q*n-Q_12-Q_23-Q_34;\n", "print (\"Q_41 = \"), (Q_41), (\"kJ/min\")\n", "\n", "dE_41 = 0-dE_12-dE_23-dE_34;\n", "print (\"dE_41 = \"), (dE_41), (\"kJ/min\")\n", "\n", "W_41 = Q_41-dE_41;\n", "print (\"W_41 = \"), (W_41), (\"kJ/min\")\n", "\n", "print \"Since sum(Q) = sum(W), \"\n", "print \"Rate of work output = -68000 KJ/min\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "dE_12 = -4340.0 kJ/min\n", "dE_23 = 42000.0 kJ/min\n", "W_34 = 69000.0 kJ/min\n", "Q_41 = -105800.0 kJ/min\n", "dE_41 = 35540.0 kJ/min\n", "W_41 = -141340.0 kJ/min\n", "Since sum(Q) = sum(W), \n", "Rate of work output = -68000 KJ/min\n" ] } ], "prompt_number": 30 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.13 page no : 128" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the steam flow round the cycle in kg/s.\n", "'''\n", "\n", "# Variables\n", "P = 1200.; \t\t\t#kW\n", "Qin = 3360.; \t\t\t#kJ/kg\n", "Qout = 2520.; \t\t\t#kJ/kg\n", "F = 6.; \t\t\t#kW\n", "\n", "# Calculations\n", "dQ = Qin - Qout;\n", "dW = P-F; \t\t\t#kJ/s\n", "m = dW/dQ;\n", "\n", "# Results\n", "print (\"Steam flow round the cycle %.3f\")%(m), (\"kg/s\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Steam flow round the cycle 1.421 kg/s\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.14 page no : 129" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) The change in internal energy ;\n", "(ii) The work done.\n", "'''\n", "\n", "# Variables\n", "dT = 25.; \t\t\t#0C\n", "Q = 30.; \t\t\t#kJ\n", "cv = 1.2; \t\t\t#kJ/kg.0C\n", "m = 2.5; \t\t\t#kg\n", "\n", "# Calculations\n", "dU = m*cv*dT;\n", "\n", "# Results\n", "print (\"change in internal energy = \"),(dU), (\"kJ\")\n", "\n", "W = Q - dU;\n", "print (\"Work done = \"),(W),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "change in internal energy = 75.0 kJ\n", "Work done = -45.0 kJ\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.15 page no : 129" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Find the change in energy of the system.\n", "(ii) The system is returned to its initial volume by an adiabatic process which requires\n", "110 kJ of work. Find the change in energy of the system.\n", "(iii) For the combined processes of (i) and (ii) determine the change in energy of the system.\n", "'''\n", "\n", "# Variables\n", "Q = 50.; \t\t\t#kJ\n", "dV = 0.14; \t\t\t#m**3\n", "p = 1.2*10**5; \t\t#N/m**2\n", "m = 90.; \t\t\t#kg\n", "d = 5.5; \t\t\t#m\n", "g = 9.8; \t\t\t#m/s**2\n", "W_adb = -110.; \t#kJ\n", "Wnet = m*g*d/1000; \t#kJ\n", "\n", "# Calculations and Results\n", "W = p*dV/1000 + Wnet; \t\t\t#kJ\n", "dE = Q-W;\n", "print (\"(i)Change in internal energy %.3f kJ\")%(dE)\n", "\n", "Q = 0;\n", "dE = -W_adb;\n", "print (\"(ii) Adiabatic process %.3f kJ\")%(dE)\n", "\n", "\n", "Q = 50.; \t\t\t#kJ\n", "dE = Q - (W_adb+W);\n", "print (\"(iii)Change in internal energy %.3f kJ\")%(dE)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)Change in internal energy 28.349 kJ\n", "(ii) Adiabatic process 110.000 kJ\n", "(iii)Change in internal energy 138.349 kJ\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.16 page no : 130" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) Change in internal energy ;\n", "(ii) Change in enthalpy.\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "V1 = 0.15; \t\t\t#m**3\n", "V2 = 0.05; \t\t\t#m**3\n", "Q = -45.; \t\t\t#kJ\n", "p1 = (5./V1+1.5)*10**5; \t\t\t#N/m**2\n", "p2 = (5./V2+1.5)*10**5; \t\t\t#N/m**2\n", "\n", "# Calculations\n", "def f0( V): \n", "\t return (5/V+1.5)*10**2\n", "\n", "W = quad(f0, V1, V2)[0]\n", "\n", "dU = Q-W;\n", "print (\"(i)Change in internal energy = %.3f kJ\")%(dU)\n", "\n", "dH = (dU*10**3+(p2*V2-p1*V1))/10**3;\n", "print (\"(ii) Change in enthalpy = %.3f kJ\")%(dH)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)Change in internal energy = 519.306 kJ\n", "(ii) Change in enthalpy = 504.306 kJ\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.17 page no : 131" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine:\n", "(i) If the expansion is quasi-static, find Q, \u2206U and W for the process.\n", "(ii) In another process, the same system expands according to the same pressure-volume\n", "relationship as in part (i), and from the same initial state to the same final state as in part (i), but\n", "the heat transfer in this case is 32 kJ. Find the work transfer for this process.\n", "(iii) Explain the difference in work transfer in parts (i) and (ii).\n", "'''\n", "\n", "# Variables\n", "V1 = 0.25; \t\t\t#m**3\n", "p1 = 500.; \t\t\t#kPa\n", "p2 = 100.; \t\t\t#kPa\n", "\n", "# Calculations and Results\n", "V2 = V1*(p1/p2)**(1/1.25)\n", "n = 1.25\n", "dU = 3.64*(p2*V2 - p1*V1)\n", "\n", "print (\"(i) If the expansion is quasi-static\")\n", "W = (p1*V1-p2*V2)/(n-1);\n", "Q = dU+W\n", "print (\"Heat transfered = %.3f\")%(Q),(\"kJ\")\n", "\n", "print (\"(ii) In another process\")\n", "Q = 32; \t\t\t#kJ\n", "W = Q-dU;\n", "print (\"Work done = %.3f\")%(W),(\"kJ\")\n", "\n", "print (\"(iii)The difference\")\n", "print (\" The work in (ii) is not equal to \u222b p dV math.since the process is not quasi-static.\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) If the expansion is quasi-static\n", "Heat transfered = 12.385 kJ\n", "(ii) In another process\n", "Work done = 157.225 kJ\n", "(iii)The difference\n", " The work in (ii) is not equal to \u222b p dV math.since the process is not quasi-static.\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.18 page no : 132" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "find\n", "(i) Heat added/kg ;\n", "(ii) Work done/kg ;\n", "(iii) Change in internal energy/kg ;\n", "(iv) Change in enthalpy/kg.\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "v1 = 0.3; \t\t\t#m**3/kg\n", "T1 = 20.; \t\t\t#0C\n", "v2 = 0.55; \t\t\t#m**3/kg\n", "T2 = 260; \t\t\t#0C\n", "p = 1.6*10**5; \t\t\t#Pa\n", "\n", "print (\"(i)Heat added per kg = \")\n", "\n", "# Calculations and Results\n", "def f5( T): \n", "\t return 1.5 + 75/(T+45)\n", "\n", "Q = quad(f5, T1,T2)[0]\n", "\n", "print (\"Q = %.3f\")%(Q), (\"kJ/kg\")\n", "\n", "\n", "print (\"(ii)The work done per kg of fluid\")\n", "W = p*(v2-v1)/1000; \t\t\t#kJ/kg\n", "print (\"W = %.3f\")%(W),(\"kJ/kg\")\n", "\n", "\n", "print (\"(iii)Change in internal energy\")\n", "dU = Q-W;\n", "print (\"dU = %.3f\")%(dU),(\"kJ/kg\")\n", "\n", "\n", "print (\"(iv)Change in enthalpy\")\n", "dH = Q;\n", "print (\"dH = %.3f\")%(dH),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)Heat added per kg = \n", "Q = 475.944 kJ/kg\n", "(ii)The work done per kg of fluid\n", "W = 40.000 kJ/kg\n", "(iii)Change in internal energy\n", "dU = 435.944 kJ/kg\n", "(iv)Change in enthalpy\n", "dH = 475.944 kJ/kg\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.19 page no : 133" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the work done during the process.\n", "'''\n", "\n", "# Variables\n", "m = 1.; \t\t\t#kg\n", "du = -42000.; \t\t\t#J\n", "cp = 840.; \t\t\t#J/kg.0C\n", "cv = 600.; \t\t\t#J/kg.0C\n", "\n", "# Calculations\n", "dT = du/m/cv;\n", "Q = m*cp*dT;\n", "W = (Q-du)/10**3;\n", "\n", "# Results\n", "print (\"Work done = \"),(W),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done = -16.8 kJ\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 4.20 page no : 133" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "find the direction and magnitude of the work and heat transfer.\n", "'''\n", "\n", "from numpy import *\n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "p1 = 190.; \t\t\t#kPa\n", "V1 = 0.035; \t\t\t#m**3\n", "p2 = 420.; \t\t\t#kPa\n", "V2 = 0.07; \t\t\t#m**3\n", "dU = 3.6*(p2*V2-p1*V1);\n", "p = [[1,0.035],[1,0.07]]\n", "q = [190,420];\n", "#X = linalg.inv(p)*q;\n", "X = linalg.solve(p,q)\n", "a = X[0]\n", "b = X[1]\n", "\n", "# Calculations\n", "def f4( V): \n", "\t return a+b*V\n", "\n", "W = quad(f4, V1, V2)[0]\n", "\n", "# Results\n", "print (\"Work done by the system = \"),(W),(\"kJ\")\n", "\n", "\n", "Q = dU+W;\n", "print (\"Heat transfer into the system = \"),(Q),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done by the system = 10.675 kJ\n", "Heat transfer into the system = 92.575 kJ\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.21 page no : 134" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) The adiabatic work ;\n", "(ii) The values of internal energy at all end states if initial value is 105 kJ.\n", "'''\n", "\n", "# Variables\n", "Qv = 90.; \t\t\t#kJ\n", "Qp = -95; \t\t\t#kJ\n", "W = -18; \t\t\t#kJ\n", "U_l = 105; \t\t\t#kJ\n", "W_lm = 0;\n", "Q_lm = 90;\n", "\n", "# Calculations\n", "U_m = U_l+90;\n", "dU_mn = Qp-W;\n", "U_n = U_m+dU_mn;\n", "dQ = Qv+Qp;\n", "dW = dQ;\n", "W_nl = dW-W;\n", "\n", "# Results\n", "print (\"W_nl(in kJ) = \"),(W_nl)\n", "\n", "print (\"U_l in kJ = \"),(U_l)\n", "\n", "print (\"U_m in kJ = \"),(U_m)\n", "print (\"U_n in kJ\"), (U_n)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "W_nl(in kJ) = 13.0\n", "U_l in kJ = 105\n", "U_m in kJ = 195\n", "U_n in kJ 118\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.23 page no : 136" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) The work done ;\n", "(ii) The index of expansion, if the above processes are replaced by a single reversible polytropic\n", "process giving the same work between the same initial and final states.\n", "'''\n", "\n", "# Variables\n", "V1 = 0.2; \t\t\t#m**3\n", "p1 = 4.*10**5; \t\t#N/m**2\n", "T1 = 403.; \t\t\t#K\n", "p2 = 1.02*10**5; \t#N/m**2\n", "dH = 72.5; \t\t\t#kJ\n", "Q_23 = dH;\n", "cp = 1.; \t\t\t#kJ/kg\n", "cv = 0.714; \t\t#kJ/kg\n", "y = 1.4;\n", "\n", "# Calculations\n", "V2 = round(V1*(p1/p2)**(1/y),2);\n", "T2 = round(T1*((p2/p1)**((y-1)/y)),1);\n", "R = (cp-cv)*1000; \t\t\t#J/kg.K\n", "m = round(p1*V1/R/T1,3);\n", "T3 = round(Q_23/(m*cp) +T2);\n", "V3 = 0.732 # round(V2*T3/T2,2);\n", "W_12 = (p1*V1 - p2*V2)/(y-1);\n", "W_23 = p2*(V3-V2);\n", "W_123 = W_12+W_23;\n", "\n", "# Results\n", "print (\"Total work done = %.3f\")%(W_123),(\"J\")\n", "\n", "print (\"(ii) Index of expansion, n\")\n", "p3 = p2;\n", "n = (p1*V1-p3*V3)/W_123 + 1;\n", "print (\"value of index = %.3f\")%(n)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Total work done = 85454.000 J\n", "(ii) Index of expansion, n\n", "value of index = 1.062\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.25 page no : 139" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Evaluate :\n", "(i) The work ;\n", "(ii) Decrease in internal energy of the system.\n", "'''\n", "\n", "# Variables\n", "d = 0.15; \t\t\t#m\n", "T = 303.; \t\t\t#K\n", "p = 3.*10**5; \t\t#N/m**2\n", "l = 0.085; \t\t\t#m\n", "Q = -4000.; \t\t#J\n", "\n", "# Calculations and Results\n", "print (\"(i) Workdone by the system\")\n", "dv = math.pi/4*d**2*l;\n", "W = p*dv;\n", "print (\"W = %.3f\")%(W/10**3),(\"kJ\")\n", "\n", "print (\"(ii) Decrease in internal energy of the system\")\n", "dU = (Q-W)/10**3;\n", "print (\"Decrease in internal energy = %.3f\")%(-dU),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Workdone by the system\n", "W = 0.451 kJ\n", "(ii) Decrease in internal energy of the system\n", "Decrease in internal energy = 4.451 kJ\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.27 page no : 140" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) The mass of gas ;\n", "(ii) The value of index \u2018n\u2019 for compression ;\n", "(iii) The increase in internal energy of the gas ;\n", "(iv) The heat received or rejected by the gas during compression.\n", "Take \u03b3 = 1.4, R = 294.2 J/kg\u00b0C.\n", "'''\n", "import math\n", "\n", "# Variables\n", "y = 1.4\n", "R = 294.2; \t\t\t#J/kg.0C\n", "p1 = 1.*10**5; \t\t#N/m**2\n", "T1 = 353.; \t\t\t#K\n", "V1 = 0.45; \t\t\t#m**3\n", "V2 = 0.13; \t\t\t#m**3\n", "p2 = 5.*10**5; \t\t#N/m**2\n", "\n", "# Calculations and Results\n", "cv = R/(y-1);\n", "print (\"(i) The mass of gas\")\n", "m = p1*V1/R/T1;\n", "print (\"m = %.3f\")%(m),(\"kg\")\n", "\n", "print (\"(ii) The value of index \u2018n\u2019 for compression\")\n", "n = math.log(p2/p1)/math.log(V1/V2);\n", "print (\"n = %.3f\")%(n)\n", "\n", "print (\"(iii) The increase in internal energy of the gas\")\n", "T2 = T1*(V1/V2)**(n-1);\n", "dU = m*cv*(T2-T1)/10**3;\n", "print (\"dU = %.3f\")%(dU),(\"kJ\")\n", "\n", "print (\"(iv) The heat received or rejected by the gas during compression.\")\n", "W = m*R*(T1-T2)/(n-1)/10**3;\n", "Q = dU+W;\n", "print (\"Q = %.3f\")%(Q),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) The mass of gas\n", "m = 0.433 kg\n", "(ii) The value of index \u2018n\u2019 for compression\n", "n = 1.296\n", "(iii) The increase in internal energy of the gas\n", "dU = 50.000 kJ\n", "(iv) The heat received or rejected by the gas during compression.\n", "Q = -17.535 kJ\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.28 page no : 141" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) The final temperature ;\n", "(ii) The final volume ;\n", "(iii) The work done.\n", "'''\n", "\n", "# Variables\n", "p1 = 1.02*10**5; \t#Pa\n", "T1 = 295.; \t\t\t#K\n", "V1 = 0.015; \t\t#m**3\n", "p2 = 6.8*10**5; \t#Pa\n", "y = 1.4;\n", "\n", "# Calculations and Results\n", "print (\"(i) Final temperature\")\n", "T2 = T1*(p2/p1)**((y-1)/y);\n", "t2 = T2-273; \n", "print (\"t2 = %.3f\")%(t2),(\"\u00b0C\")\n", "\n", "\n", "print (\"(ii) Final volume :\")\n", "V2 = V1*(p1/p2)**(1/y);\n", "print (\"V2 = %.3f\")%(V2),(\"m**3\")\n", "\n", "\n", "print (\"(iii)Work done\")\n", "R = 287;\n", "m = p1*V1/R/T1;\n", "W = m*R*(T1-T2)/(y-1)/10**3;\n", "print (\"W = %.3f\")%(W),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Final temperature\n", "t2 = 234.253 \u00b0C\n", "(ii) Final volume :\n", "V2 = 0.004 m**3\n", "(iii)Work done\n", "W = -2.752 kJ\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.29 page no : 142" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate c p and c v .\n", "'''\n", "\n", "from numpy import *\n", "import math\n", "\n", "# Variables\n", "m = 0.44; \t\t\t#kg\n", "T1 = 453.; \t\t\t#K\n", "ratio = 3.; \t\t#ratio = V2/V1\n", "T2 = 288.; \t\t\t#K\n", "W_12 = 52.5; \t\t#kJ\n", "\n", "# Calculations\n", "y = math.log(T2/T1)/ math.log(1/ratio) + 1;\n", "R = W_12*(y-1)/m/(T1-T2);\n", "M = [[1,-1],[1,-y]];\n", "N = [R,0];\n", "X = linalg.inv(M)*N;\n", "cp = X[0][0];\n", "cv = X[1][0];\n", "\n", "# Results\n", "print (\"cp = %.3f\")%(cp),(\"kJ/kg.K\")\n", "\n", "print (\"cv = %.3f\")%(cv),(\"kJ/kg.K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "cp = 1.021 kJ/kg.K\n", "cv = 0.723 kJ/kg.K\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.30 page no : 143" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the heat flow to or from the cylinder walls.\n", "'''\n", "\n", "# Variables\n", "n = 1.3;\n", "m = 1; \t\t\t#kg\n", "p1 = 1.1; \t\t\t#bar\n", "T1 = 300.; \t\t\t#K\n", "p2 = 6.6; \t\t\t#bar\n", "R0 = 8314.;\n", "M = 30.;\n", "cp = 1.75; \t\t\t#kJ/kg.K\n", "\n", "\n", "# Calculations\n", "R = R0/M/1000; \t\t\t#kJ/kg.K\n", "cv = cp - R;\n", "y = cp/cv;\n", "T2 = T1 *(p2/p1)**((n-1)/n);\n", "W = R*(T1-T2)/(n-1);\n", "Q = ((y-n)/(y-1))*W;\n", "\n", "# Results\n", "print (\"Heat supplied = %.3f\")%(Q),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat supplied = 84.352 kJ/kg\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.31 page no : 144" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) Pressure at the end of constant volume cooling.\n", "(ii) Change in internal energy during constant volume process.\n", "(iii) Net work done and heat transferred during the cycle. Assume\n", "'''\n", "import math\n", "\n", "# Variables\n", "cp = 14.3; \t\t\t#kJ/kg.K\n", "cv = 10.2; \t\t\t#kJ/kg.K\n", "V1 = 0.1; \t\t\t#m**3\n", "T1 = 300.; \t\t\t#K\n", "p1 = 1.; \t\t\t#bar\n", "p2 = 8.; \t\t\t#bar\n", "y = cp/cv;\n", "R = cp-cv;\n", "V2 = V1*(p1/p2)**(1/y);\n", "V3 = V2;\n", "T2 = T1*(p2/p1)**((y-1)/y);\n", "p3 = p1*V1/V3;\n", "T3 = 300.; \t\t\t#K\n", "\n", "\n", "# Calculations and Results\n", "print (\"(i) Pressure at the end of consmath.tant volume cooling = %.3f\")%(p3),(\"bar\")\n", "\n", "print (\"(ii) Change in internal energy during consmath.tant volume process\")\n", "m = p1*V1/R/T1*10**2; \t\t\t#kg\n", "\n", "dU_23 = m*cv*(T3-T2);\n", "print (\"dU_23 = %.3f\")%(dU_23),(\"kJ\")\n", "\n", "print (\"(iii) Net work done and heat transferred during the cycle\")\n", "W_12 = m*R*(T1-T2)/(y-1);\n", "W_23 = 0;\n", "W_31 = p3*V3*math.log(V1/V3)*10**2; \t\t\t#kJ\n", "Wnet = W_12+W_23+W_31;\n", "print (\"Net work done = %.3f\")%(Wnet),(\"kJ\")\n", "Qnet = Wnet;\n", "print (\"Heat transferred during the complete cycle = %.3f\")%(Qnet),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Pressure at the end of consmath.tant volume cooling = 4.407 bar\n", "(ii) Change in internal energy during consmath.tant volume process\n", "dU_23 = -20.281 kJ\n", "(iii) Net work done and heat transferred during the cycle\n", "Net work done = -5.449 kJ\n", "Heat transferred during the complete cycle = -5.449 kJ\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.32 page no : 145" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the net work done and heat transferred during the cycle.\n", "'''\n", "\n", "import math\n", "\n", "# Variables\n", "V1 = 0.15; \t\t\t#m**3\n", "p1 = 15.; \t\t\t#bar\n", "T1 = 550.; \t\t\t#K\n", "T2 = T1;\n", "r = 4.; \t\t\t#r = V2/V1\n", "V2 = r*V1;\n", "T3 = 290.; \t\t\t#K\n", "\n", "# Calculations\n", "p2 = p1*V1/V2;\n", "W_12 = p1*V1*math.log(V2/V1)*10**2; \t\t\t#kJ\n", "V3 = V2;\n", "p3 = p2*T3/T2;\n", "W_23 = 0;\n", "n = math.log(p1/p3)/math.log(V3/V1);\n", "W_31 = (p3*V3-p1*V1)/(n-1)*10**2; \t\t\t#kJ\n", "\n", "# Results\n", "\n", "Wnet = W_12+W_23+W_31\n", "print (\"net work done = %.3f\")%Wnet , (\"kJ\")\n", "\n", "Qnet = Wnet;\n", "print (\"Heat transferred during the cycle = %.3f\")%(Qnet),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "net work done = 81.537 kJ\n", "Heat transferred during the cycle = 81.537 kJ\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.33 page no : 146" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Sketch the cycle on p-V diagram.\n", "(ii) Calculate the work done in each process starting whether it is done on or by the system\n", "and evaluate the net cyclic work and heat transfer.\n", "'''\n", "\n", "%pylab inline\n", "\n", "import math \n", "from scipy.integrate import quad \n", "from matplotlib.pyplot import *\n", "from numpy import *\n", "\n", "# Variables\n", "m = 1; \t\t\t#kg\n", "p1 = 5; \t\t\t#bar\n", "V1 = 0.02; \t\t\t#m**3\n", "V2 = 0.08; \t\t\t#m**3\n", "p2 = 1.5; \t\t\t#bar\n", "\n", "\n", "# Calculations and Results\n", "def f(V):\n", " return a+b*V;\n", "\n", "A = [[1,0.02],[1,0.08]]\n", "B = [5,1.5];\n", "#X = linalg.inv(A)*B;\n", "X = linalg.solve(A,B)\n", "a = X[0]\n", "b = X[1]\n", "\n", "print (\"(i) p-V diagram\")\n", "\n", "V = linspace(0.02,0.08,80);\n", "p = a+b*V;\n", "plot(V,p,'b')\n", "\n", "V = [0.0667 ,0.08];\n", "p = [1.5 ,1.5];\n", "plot(V,p,'g')\n", "\n", "V = linspace(0.02,0.0667,447)\n", "def fa(V):\n", " return 0.1/V;\n", "f = fa(V)\n", "plot(V,f,'r')\n", "suptitle(\"p-V diagram\")\n", "\n", "V = [0.0667, 0.0667];\n", "p = [1.5, 0];\n", "plot(V,p,'--')\n", "xlabel(\"V(m)**3\")\n", "ylabel(\"P(bar)\")\n", "text(.04,4,'Reversible Expansion')\n", "text(0.04,2.3,\"pV = C\")\n", "text(0.07,1.2,\"p = C\")\n", "\n", "print (\"(ii) Work done and heat transfer\")\n", "\n", "\n", "def f7(V): \n", "\t return (a+b*V)*10**2\n", "\n", "W_12 = quad(f7,V1,V2)[0]\n", "\n", "print (\"Work done by the system = \"),(W_12), (\"kJ\")\n", "\n", "p3 = p2;\n", "V3 = round(p1*V1/p3,4);\n", "W_23 = p2*(V3-V2)*10**2; \t\t\t#kJ\n", "\n", "W_31 = round(p3*V3*math.log(V1/V3)*10**2,2); \t\t\t#kJ\n", "print (\"Work done on the system = %.3f\")% (W_31), (\"kJ\")\n", "\n", "W_net = W_12+W_23+W_31;\n", "print (\"Net work done = %.3f\")% (W_net), (\"kJ\")\n", "\n", "Q_net = W_net;\n", "print (\"Heat transferred during the complete cycle = %.3f\")% (Q_net),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n", "(i) p-V diagram" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "\n", "(ii) Work done and heat transfer" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "\n", "Work done by the system = 19.5 kJ\n", "Work done on the system = -12.050 kJ\n", "Net work done = 5.455 kJ\n", "Heat transferred during the complete cycle = 5.455 kJ\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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9S/fKsZhl2EcIgf/85z8YM2YM/vnPf+LYsWNGNUh2oGZN4O235Yrj3bsDffvK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\nLTh8+DBWr16NtLQ0pZqzSkl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"text": [ "" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.34 page no : 147" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine theamount of heat added to the system so that piston moves by 3.5 cm.\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "\n", "# Variables\n", "cv = 0.71; \t\t\t#kJ/kg.K\n", "R = 0.287; \t\t\t#kJ/kg.K\n", "d = 8.; \t\t\t#cm\n", "l = 3.5; \t\t\t#cm\n", "S = 150.; \t\t\t#N/cm\n", "p1 = 30.; \t\t\t#N/cm\n", "V1 = 45.; \t\t\t#cm**3\n", "T1 = 293.; \t\t\t#K\n", "cv = 0.71; \t\t\t#kJ/kg.K\n", "R = 0.287; \t\t\t#kJ/kg.K\n", "\n", "# Calculations\n", "A = math.pi/4*d**2;\n", "C = p1-S/A**2*V1;\n", "dV = l*A;\n", "V2 = V1+dV;\n", "p2 = S/A**2*V2 + C;\n", "\n", "def f3( p): \n", "\t return A**2/S*p/100\n", "\n", "W = quad(f3, p1, p2)[0]\n", "\n", "T2 = p2*V2*T1/p1/V1;\n", "m = p1*V1/R/T1/10**5; \t\t\t#kg\n", "dU = m*cv*(T2-T1);\n", "Q_12 = dU + W*10**(-3);\n", "\n", "# Results\n", "print (\"Amount of heat added to the system = %.3f\")% (Q_12), (\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Amount of heat added to the system = 0.250 kJ\n" ] } ], "prompt_number": 31 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.35 page no : 164" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) The change in enthalpy (\u2206 h) ;\n", "(ii) Work done during the process (W).\n", "'''\n", "\n", "# Variables\n", "r = 10.; \t\t\t#kg/min\n", "p1 = 1.5*10**5; \t\t\t#N/m**2\n", "rho1 = 26.; \t\t\t#kg/m**3\n", "C1 = 110.; \t\t\t#m/s\n", "u1 = 910.; \t\t\t#kJ/kg\n", "p2 = 5.5*10**5; \t\t\t#N/m**2\n", "rho2 = 5.5; \t\t\t#kg/m**3\n", "C2 = 190.; \t\t\t#m/s\n", "u2 = 710.; \t\t\t#kJ/kg\n", "Q = 55.; \t\t\t#kJ/s\n", "h = 55.; \t\t\t#m\n", "g = 9.81; \t\t\t#m/s**2\n", "v2 = 1/rho2;\n", "v1 = 1/rho1;\n", "\n", "# Calculations and Results\n", "\n", "dh = u2-u1+ (p2*v2-p1*v1)/10**3;\n", "print (\"(i) Change in enthalpy %.3f\")%(dh), (\"kJ/kg\")\n", "\n", "print (\"(ii) Work done during the process (W).\")\n", "\n", "Q = 330.; \t\t\t#kJ/kg\n", "KE = (C2**2-C1**2)/2/10**3; \t\t\t#kJ\n", "PE = g*h/10**3; \t\t\t#kJ\n", "W = -Q-KE-PE-dh;\n", "print (\"Work done = %.3f\")%(W),(\"kJ\")\n", "\n", "\n", "P = W*10/60.;\n", "print (\"Work done per second = %.3f\")%(P),(\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Change in enthalpy -105.769 kJ/kg\n", "(ii) Work done during the process (W).\n", "Work done = -236.770 kJ\n", "Work done per second = -39.462 kW\n" ] } ], "prompt_number": 32 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.36 page no : 166" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) The rate at which heat is rejected to the turbine, and\n", "(ii) The area of the inlet pipe given that the specific volume of the gases at the inlet is\n", "0.45 m 3 /kg.\n", "'''\n", "\n", "# Variables\n", "\n", "import math \n", "\n", "m = 15.; \t\t\t#kg/s\n", "v = 0.45; \t\t\t#m**3/kg\n", "P = 12000.; \t\t#kW\n", "W = P/m; \t\t\t#kJ/kg\n", "h1 = 1260.; \t\t#kJ/kg\n", "h2 = 400.; \t\t\t#kJ/kg\n", "C1 = 50.; \t\t\t#m/s\n", "C2 = 110.; \t\t\t#m/s\n", "\n", "# Calculations and Results\n", "print (\"(i) Heat rejected = \"),\n", "Q = h2-h1+(C2**2-C1**2)/2/10**3 +W;\n", "Qnet = m*Q;\n", "print (\"Qnet = %.3f\")%(-Qnet),(\"kW\")\n", "\n", "print (\"(ii) Inlet area\")\n", "A = v*m/C1;\n", "print (\"A = %.3f\")%(A),(\"m**2\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Heat rejected = Qnet = 828.000 kW\n", "(ii) Inlet area\n", "A = 0.135 m**2\n" ] } ], "prompt_number": 33 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.37 page no : 167" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) The power required to drive the compressor ;\n", "(ii) The inlet and output pipe cross-sectional areas.\n", "'''\n", "\n", "# Variables\n", "m = 0.5; \t\t\t#kg/s\n", "C1 = 6.; \t\t\t#m/s\n", "C2 = 5.; \t\t\t#m/s\n", "p1 = 1.; \t\t\t#bar\n", "p2 = 7.; \t\t\t#bar\n", "v1 = 0.85; \t\t\t#m**3/kg\n", "v2 = 0.16; \t\t\t#m**3/kg\n", "du = 90.; \t\t\t#kJ/kg\n", "Q = -120.; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "print (\"(i) Power required to drive the compressor\")\n", "W = -du+(C1**2-C2**2)/2/1000 + (p1*v1 - p2*v2)*10**2 + Q;\n", "Power = m*W; \n", "print (\"Power = %.3f\")%(-Power),(\"kW\")\n", "\n", "\n", "print (\"(ii) Inlet and outlet pipe cross-sectional areas\")\n", "A1 = m*v1/C1;\n", "A2 = m*v2/C2;\n", "print (\"Inlet crosssectional area = %.3f\")% (A1), (\"m**2\")\n", "\n", "print (\"Outlet crossectional area = %.3f\")%(A2), (\"m**2\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Power required to drive the compressor\n", "Power = 118.497 kW\n", "(ii) Inlet and outlet pipe cross-sectional areas\n", "Inlet crosssectional area = 0.071 m**2\n", "Outlet crossectional area = 0.016 m**2\n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.38 page no : 168" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "calculate the power developed by the turbine. Consider the boiler and turbine as single system\n", "'''\n", "\n", "# Variables\n", "h1 = 800.; \t\t\t#kJ/kg\n", "C1 = 5.; \t\t\t#m/s\n", "h2 = 2520.; \t\t\t#kJ/kg\n", "C2 = 50.; \t\t\t#m/s\n", "dZ = 4.; \t\t\t#m\n", "g = 9.81; \t\t\t#m/s**2\n", "Q = 2180.; \t\t\t#kJ/kg\n", "\n", "# Calculations\n", "W = h1-h2+(C1**2 - C2**2)/2/1000 +dZ*g/1000+Q;\n", "\n", "# Results\n", "print (\"Power developed = %.3f\")%(W), (\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Power developed = 458.802 kW\n" ] } ], "prompt_number": 35 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.39 page no : 169" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the power output of the turbine.\n", "'''\n", "\n", "# Variables\n", "g = 9.8; \t\t\t#m/s**2\n", "m = 4500./3600; \t#kg/s\n", "C1 = 2800./60; \t\t#m/s\n", "Z1 = 5.5; \t\t\t#m\n", "h1 = 2800.; \t\t#kJ/g\n", "C2 = 5600./60; \t\t#m/s\n", "Z2 = 1.5; \t\t\t#m\n", "h2 = 2300.; \t\t#kJ/kg\n", "Q = -16000./3600; \t#kJ/s\n", "\n", "# Calculations\n", "W = Q-m*((h1-h2) + (C2**2 - C1**2)/2/1000 + (Z2-Z1)*g/1000);\n", "\n", "# Results\n", "print (\"Power output of the turbine = %.3f\")% (-W),(\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Power output of the turbine = 633.479 kW\n" ] } ], "prompt_number": 36 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.40 page no : 170" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the final enthalpy of steam.\n", "'''\n", "\n", "# Variables\n", "p1 = 6.87; \t\t\t#bar\n", "C1 = 50.; \t\t\t#m/s\n", "p2 = 1.37; \t\t\t#bar\n", "C2 = 500.; \t\t\t#m/s\n", "print (\"From steam table corresponding to p1\")\n", "h1 = 2850.; \t\t\t#kJ/kg\n", "\n", "# Calculations\n", "h2 = h1 - (C2**2-C1**2)/2/1000;\n", "\n", "# Results\n", "print (\"Final enthalpy of steam = \"), (h2),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From steam table corresponding to p1\n", "Final enthalpy of steam = 2726.25 kJ\n" ] } ], "prompt_number": 37 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.41 page no : 171" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the power capacity of the system in MW.\n", "'''\n", "\n", "# Variables\n", "m = 220./60; \t\t\t#kg/s\n", "C1 = 320.; \t\t\t#m/s\n", "p1 = 6*10.**5; \t\t\t#N/m**2\n", "u1 = 2000.*10**3; \t\t\t#J/kg\n", "v1 = 0.36; \t\t\t#m**3/kg\n", "C2 = 140.; \t\t\t#m/s\n", "p2 = 1.2*10**5; \t\t\t#N/m**2\n", "u2 = 1400.*10**3; \t\t\t#J/kg\n", "v2 = 1.3; \t\t\t#m**3/kg\n", "Q = 100*10.**3; \t\t\t#J/s\n", "\n", "# Calculations\n", "W = (m*((u1-u2)+ (p1*v1 - p2*v2) + (C1**2-C2**2)/2) -Q)/10**6;\n", "\n", "# Results\n", "print (\"power capacity of the system = %.3f\")% (W),(\"MW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "power capacity of the system = 2.472 MW\n" ] } ], "prompt_number": 38 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.42 page no : 172" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the capacity of the turbine if the gas flow is 5 kg/s.\n", "'''\n", "\n", "\n", "# Variables\n", "p1 = 7.5*10**5; \t\t\t#N/m**2\n", "C1 = 140.; \t\t\t#m/s\n", "h1 = 950.*10**3; \t\t\t#J/kg\n", "p2 = 2*10.**5; \t\t\t#N/m**2\n", "C2 = 280.; \t\t\t#m/s\n", "h2 = 650.*10**3; \t\t\t#J/kg\n", "m = 5.; \t\t\t#kg/s\n", "\n", "# Calculations\n", "W = (h1-h2)+(C1**2-C2**2)/2\n", "Power = m*W/1000;\n", "\n", "# Results\n", "print (\"Power capacity of turbine = \"), (Power), (\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Power capacity of turbine = 1353.0 kW\n" ] } ], "prompt_number": 39 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.43 page no : 173" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Find : (i) Motor power required to drive the compressor ;\n", "(ii) Ratio of inlet to outlet pipe diameter.\n", "'''\n", "\n", "# Variables\n", "C1 = 12.; \t\t\t#m/s\n", "p1 = 1.*10**5; \t\t\t#N/m**2\n", "v1 = 0.5; \t\t\t#m**3/kg\n", "C2 = 90.; \t\t\t#m/s\n", "p2 = 8.*10**5; \t\t\t#N/m**2\n", "v2 = 0.14; \t\t\t#m**3/kg\n", "dh = 150.; \t\t\t#kJ/kg\n", "Q = -11.67; \t\t\t#kJ/s\n", "m = 0.2; \t\t\t#kg/s\n", "\n", "# Calculations and Results\n", "print (\"(i) Motor power required to drive the compressor\")\n", "W = m*(-dh + (C1**2-C2**2)/2/1000) +Q;\n", "print (\"Power = %.3f\")% (-W), (\"kW\")\n", "\n", "\n", "print (\"(ii)Ratio of inlet to outlet pipi diameter\")\n", "ratio = math.sqrt(C2/C1*v1/v2);\n", "print (\"ratio = %.3f\")% (ratio)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Motor power required to drive the compressor\n", "Power = 42.466 kW\n", "(ii)Ratio of inlet to outlet pipi diameter\n", "ratio = 5.175\n" ] } ], "prompt_number": 40 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.44 page no : 175" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Compute the amount of heat transfer to the atmosphere from the compressor per kg of air.\n", "'''\n", "\n", "# Variables\n", "W = -175.; \t\t\t #kJ/kg\n", "dh = 70.; \t\t \t#kJ/kg\n", "Q_water = -92.; \t\t\t#kJ/kg\n", "\n", "# Calculations\n", "Q = dh+W;\n", "Q_atm = Q-Q_water;\n", "\n", "# Results\n", "print (\"Heat transferred to the atmosphere = \"),(-Q_atm), (\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat transferred to the atmosphere = 13.0 kJ/kg\n" ] } ], "prompt_number": 41 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.45 page no : 176" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "'''\n", "(i) Find the velocity at exit of the nozzle.\n", "(ii) If the inlet area is 900 cm 2 and the specific volume at inlet is 0.187 m 3 /kg, find the mass\n", "flow rate.\n", "(iii) If the specific volume at the nozzle exit is 0.498 m 3 /kg, find the exit area of nozzle.\n", "'''\n", "import math\n", "\n", "# Variables\n", "h1 = 2800.*10**3; \t\t\t#J/kg\n", "C1 = 50.; \t\t\t#m/s\n", "A1 = 900.*10**(-4); \t\t\t#m**2\n", "v1 = 0.187; \t\t\t#m**3/kg\n", "h2 = 2600.*10**3; \t\t\t#J/kg\n", "v2 = 0.498; \t\t\t#m**3/kJ\n", "\n", "# Calculations and Results\n", "print (\"(i) Velocity at exit of the nozzle\")\n", "C2 = math.sqrt(2*((h1-h2) + C1**2/2));\n", "\n", "print (\"C2 = %.3f\")% (C2),(\"m/s\")\n", "\n", "\n", "print (\"(ii) Mass flow rate\")\n", "m = A1*C1/v1;\n", "print (\"m = %.3f\")% (m), (\"kg/s\")\n", "\n", "\n", "print (\"(iii) Area at the exit\")\n", "A2 = m*v2/C2*10**4;\n", "print (\"A2 = %.3f\")%(A2), (\"cm**2\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Velocity at exit of the nozzle\n", "C2 = 634.429 m/s\n", "(ii) Mass flow rate\n", "m = 24.064 kg/s\n", "(iii) Area at the exit\n", "A2 = 188.894 cm**2\n" ] } ], "prompt_number": 42 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.46 page no : 177" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "evaluate the heat transfer from the water per kg of water flowing.\n", "'''\n", "\n", "# Variables\n", "h1 = 240.; \t\t\t#kJ/kg\n", "h2 = 192.; \t\t\t#kJ/kg\n", "dZ = 20.; \t\t\t#m\n", "g = 9.81; \t\t\t#m/s**2\n", "\n", "# Calculations\n", "Q = (h2-h1)+dZ*g/1000;\n", "\n", "# Results\n", "print (\"heat transfer = %.3f\")% (-Q), (\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "heat transfer = 47.804 kJ/kg\n" ] } ], "prompt_number": 43 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.47 page no : 178" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "evaluate the relative velocity of gas leaving the jet pipe. For the gas at t = 820\u00b0C, h = 800 kJ/kg and at 910\u00b0C, 915 kJ/kg.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "p1 = 2.; \t\t\t#bar\n", "C1 = 300.; \t\t\t#m/s\n", "Q = 0.;\n", "h1 = 915.*10**3; \t\t\t#J/kg\n", "h2 = 800.*10**3; \t\t\t#J/kg\n", "\n", "# Calculations\n", "C2 = math.sqrt(2*(h1-h2 + C1**2/2));\n", "\n", "# Results\n", "print (\"Relative velocity of gas leaving the pipe = %.3f\")% (C2), (\"m/s\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Relative velocity of gas leaving the pipe = 565.685 m/s\n" ] } ], "prompt_number": 44 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.48 page no : 179" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the capacity of the electric motor to run the pump.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "mw = 50; \t\t\t#kg/s\n", "p1 = 10.**5; \t\t\t#N/m**2\n", "p2 = 4.2*10**5; \t\t\t#N/m**2\n", "h = 10.7; \t\t\t#m\n", "d1 = 0.2; \t\t\t#m\n", "d2 = 0.1; \t\t\t#m\n", "v1 = 1./1000;\n", "v2 = 1./1000;\n", "g = 9.81; \t\t\t#m/s**2\n", "\n", "# Calculations\n", "C1 = mw*4/math.pi/d1**2*v1;\n", "C2 = mw*4/math.pi/d2**2*v2;\n", "W = mw*((p1*v1-p2*v2) + (g*(0-h))+(C1**2-C2**2)/2)/10**3;\n", "\n", "# Results\n", "print (\"Capacity of electric motor %.3f\")%(-W), (\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Capacity of electric motor 22.198 kW\n" ] } ], "prompt_number": 45 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 4.49 page no : 180" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the velocity of the exhaust jet.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "Ca = 250.; \t\t\t#m/s\n", "t = -14.; \t\t\t#0C\n", "ha = 250.; \t\t\t#kJ/kg\n", "hg = 900.; \t\t\t#kJ/kg\n", "ratio = 0.0180;\n", "Ef = 45.*10**3; \t\t\t#kJ/kg\n", "Q = -21.; \t\t\t#kJ/kg\n", "ma = 1.; \t\t\t#kg\n", "mg = 1.018; \t\t\t#kg\n", "mf = 0.018; \t\t\t#kg\n", "\n", "#Calculations\n", "Eg = 0.06*mf/mg*Ef;\n", "Cg = math.sqrt(2000*((ma*(ha+Ca**2/2/1000) + mf*Ef + Q)/mg -hg-Eg));\n", "\n", "# Results\n", "print (\"velocity of exhaust gas jet = %.3f\")%(Cg),(\"m/s\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "velocity of exhaust gas jet = 455.160 m/s\n" ] } ], "prompt_number": 46 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.50 page no : 181" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "calculate :\n", "(i) Rate of heat transfer to the air in the heat exchanger ;\n", "(ii) The power output from the turbine assuming no heat loss ;\n", "'''\n", "\n", "# Variables\n", "t1 = 20.; \t\t\t#0C\n", "C1 = 40.; \t\t\t#m/s\n", "t2 = 820.; \t\t\t#0C\n", "C2 = 40.; \t\t\t#m/s\n", "t3 = 620.; \t\t\t#0C\n", "C3 = 55.; \t\t\t#m/s\n", "t4 = 510.; \t\t\t#0C\n", "m = 2.5; \t\t\t#kg/s\n", "cp = 1.005; \t\t\t#kJ/kg.0C\n", "\n", "# Calculations and Results\n", "print (\"(i) Heat exchanger\")\n", "Q_12 = m*cp*(t2-t1);\n", "print (\"rate of heat transfer = \"),(Q_12), (\"kJ/s\")\n", "\n", "print (\"(ii) Turbine\")\n", "W_23 = m*((cp*(t2-t3))+(C2**2-C3**2)/2/1000);\n", "print (\"Power output of turbine = %.3f\")%(W_23), (\"kW\")\n", "\n", "print (\"(iii) Nozzle\")\n", "C4 = math.sqrt(2*1000*(cp*(t3-t4)+C3**2/2/1000));\n", "print (\"Velocity at exit from the nozzle = %.3f\")%(C4), (\"m/s\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Heat exchanger\n", "rate of heat transfer = 2010.0 kJ/s\n", "(ii) Turbine\n", "Power output of turbine = 500.719 kW\n", "(iii) Nozzle\n", "Velocity at exit from the nozzle = 473.418 m/s\n" ] } ], "prompt_number": 47 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.51 page no : 185" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) The state of steam after cooling ;\n", "(ii) The amount of heat rejected by the steam.\n", "'''\n", "\n", "# Variables\n", "V = 0.028; \t\t\t#m**3\n", "p1 = 80.; \t\t\t#bar\n", "t = 350.; \t\t\t#0C\n", "p2 = 50.; \t\t\t#bar\n", "v1 = 0.02995; \t\t\t#m**3/kg\n", "h1 = 2987.3; \t\t\t#kJ/kg\n", "v2 = 0.02995; \t\t\t#m**3/kg\n", "vg2 = 0.0394; \t\t\t#m**3/kg\n", "uf2 = 1149.; \t\t\t#kJ/kg\n", "ug2 = 2597.; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "m = V/v1;\n", "u1 = h1 - (p1*v1*10**2); \t\t\t#kJ/kg\n", "\n", "print (\"(i) State of steam after cooling\")\n", "x2 = v2/vg2;\n", "print (\"dryness fraction = %.3f\")%(x2)\n", "\n", "\n", "print (\"(ii) Heat rejected by the steam\")\n", "u2 = (1-x2)*uf2 + x2*ug2;\n", "Q = m*(u2-u1);\n", "print (\"Heat rejected = %.3f\")% (-Q), (\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) State of steam after cooling\n", "dryness fraction = 0.760\n", "(ii) Heat rejected by the steam\n", "Heat rejected = 465.575 kJ\n" ] } ], "prompt_number": 48 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.52 page no : 188" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) Heat supplied ;\n", "(ii) Work done\n", "'''\n", "\n", "# Variables\n", "m = 0.08; \t\t\t#kg\n", "p = 2.*10**5; \t\t\t#Pa\n", "V = 0.10528; \t\t\t#m**3\n", "h1 = 2706.3; \t\t\t#kJ/kg\n", "h2 = 3071.8; \t\t\t#kJ/kg\n", "v1 = 0.885; \t\t\t#m**3/kg\n", "\n", "# Calculations and Results\n", "v2 = V/m; \t\t\t#m**3/kg\n", "\n", "print (\"(i) Heat supplied\")\n", "Q = m*(h2-h1);\n", "print (\"Q = \"),(Q), (\"kJ\")\n", "\n", "\n", "W = p*(v2-v1);\n", "W_total = m*W/10**3;\n", "print (\"(ii)Total work done = \"), (W_total), (\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Heat supplied\n", "Q = 29.24 kJ\n", "(ii)Total work done = 6.896 kJ\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.53 page no : 189" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the heat supplied, and show on a T-s diagram the area which represents the heat flow.\n", "'''\n", "from numpy import *\n", "from matplotlib.pyplot import *\n", "\n", "# Variables\n", "m = 1.; \t\t\t#kg\n", "p = 8.; \t\t\t#bar\n", "s1 = 6.55; \t\t\t#kJ/kg.K\n", "T = 200.; \t\t\t#0C\n", "s_f1 = 2.0457; \t\t\t#kJ/kg.K\n", "s_fg1 = 4.6139; \t\t\t#kJ/kg.K\n", "h_f1 = 720.9; \t\t\t#kJ/kg\n", "h_fg1 = 2046.5; \t\t\t#kJ/kg\n", "h2 = 2839.3; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "x1 = (s1-s_f1)/s_fg1;\n", "h1 = h_f1+x1*h_fg1;\n", "Q = h2-h1;\n", "print (\"Heat supplied = %.3f\")%(Q), (\"kJ/kg\")\n", "\n", "# For T-s diagram\n", "\n", "s = linspace(0,.10,10);\n", "T = (-(s-5)**2+298);\n", "plot(s,T)\n", "\n", "T = [295.44 ,295.44];\n", "s = [6.6 ,3.45];\n", "plot(s,T,'g')\n", "\n", "s = [6.6 ,7];\n", "T = [295.44, 300];\n", "plot(s,T,'g')\n", "\n", "s = [6.55 ,6.55];\n", "T = [270 ,295.44];\n", "plot(s,T,'r')\n", "\n", "s = [6.6, 6.6];\n", "T = [270 ,295.44];\n", "plot(s,T,'--r')\n", "\n", "s = [6.66, 6.66];\n", "T = [270, 295.44];\n", "plot(s,T,'r')\n", "\n", "\t\t\t#The area in red represents the heat flow and it goes upto x-axis\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat supplied = 120.513 kJ/kg\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 50, "text": [ "[]" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 50 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.54 page no : 192" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate per kg :\n", "(i) The change of internal energy ;\n", "(ii) The change of enthalpy ;\n", "(iii) The work done.\n", "'''\n", "\n", "# Variables\n", "p1 = 7.*10**5; \t\t\t#Pa\n", "p2 = 1.5*10**5; \t\t\t#Pa\n", "Q = 420.; \t\t\t#kJ/kg\n", "uf = 696.; \t\t\t#kJ/kg\n", "x = 0.95;\n", "ug = 2573.; \t\t\t#kJ/kg\n", "u_f2 = 2580.; \t\t\t#kJ/kg\n", "u_g2 = 2856.; \t\t\t#kJ/kg\n", "x2 = 15./50;\n", "h_f1 = 697.1; \t\t\t#kJ/kg\n", "h_fg1 = 2064.9; \t\t\t#kJ.kg\n", "h_f2 = 2772.6; \t\t\t#kJ/kg\n", "h_g2 = 2872.9; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "print (\"(i) Change of internal energy\")\n", "u1 = (1-x)*uf + x*ug;\n", "u2 = 2602.8; \t\t\t#kJ/kg\n", "du = u2-u1;\n", "print (\"du = \"),(du), (\"kJ/kg\")\n", "\n", "print (\"(ii) Change in enthalpy\")\n", "h1 = h_f1+x*h_fg1;\n", "h2 = h_f2+x2*(h_g2-h_f2);\n", "dh = h2-h1;\n", "print (\"dh = \"), (dh), (\"kJ/kg\")\n", "\n", "print (\"(iii) Work done \")\n", "W = Q-du;\n", "print (\"W = \"), (W), (\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Change of internal energy\n", "du = 123.65 kJ/kg\n", "(ii) Change in enthalpy\n", "dh = 143.935 kJ/kg\n", "(iii) Work done \n", "W = 296.35 kJ/kg\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.55 page no : 194" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "calculate per kg of steam :\n", "(i) Work done ;\n", "(ii) Heat flow to or from the cylinder walls.\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "\n", "p1 = 5.5*10**5; \t\t\t#Pa\n", "x1 = 1.;\n", "p2 = 0.75*10**5; \t\t\t#Pa\n", "v1 = 0.3427; \t\t\t#m**3/kg\n", "v2 = p1*v1/p2;\n", "\n", "# Since v2 > vg (at 0.75 bar), therefore, the steam is superheated at state 2.\n", "u2 = 2567.25; \t\t\t#kJ/kg\n", "u1 = 2565.; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "du = u2-u1; \t\t\t#kJ/kg\n", "C = p1*v1;\n", "\n", "print (\"Work done \"),\n", "\n", "def f6( v): \n", "\t return C/v\n", "\n", "W = quad(f6, v1,v2)[0]\n", "\n", "print (\"W =\"),(W), (\"N-m/kg\")\n", "\n", "\n", "\n", "Q = du+W/10**3;\n", "\n", "print (\"Heat supplied = %.3f\")%(Q),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done W = 375543.199592 N-m/kg\n", "Heat supplied = 377.793 kJ/kg\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.56 page no : 195" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate per kg of steam :\n", "(i) The heat supplied ;\n", "(ii) The work done.\n", "'''\n", "\n", "# Variables\n", "p1 = 100.; \t\t\t#bar\n", "p2 = 10.; \t\t\t#bar\n", "s1 = 5.619; \t\t\t#kJ/kg.K\n", "T = 584.; \t\t\t#K\n", "s2 = 7.163; \t\t\t#kJ/kg.K\n", "u1 = 2545.; \t\t\t#kJ/kg\n", "u2 = 2811.8; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "print (\"(i)Heat supplied \")\n", "Q = T*(s2-s1);\n", "print (\"Q = \"),(Q),(\"kJ/kg\")\n", "\n", "print (\"(ii) Work done\")\n", "W = Q-(u2-u1);\n", "print (\"W = \"), (W), (\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)Heat supplied \n", "Q = 901.696 kJ/kg\n", "(ii) Work done\n", "W = 634.896 kJ/kg\n" ] } ], "prompt_number": 53 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.57 page no : 198" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the work done by the steam.\n", "'''\n", "\n", "# Variables\n", "m = 1.; \t\t\t#kg\n", "p1 = 120.*10**5; \t#N/m**2\n", "t1 = 400.; \t\t\t#0C\n", "p2 = 38.; \t\t\t#bar\n", "h1 = 3051.3; \t\t#kJ/kg\n", "v1 = 0.02108; \t\t#m**3/kg\n", "\n", "# Calculations\n", "u1 = h1-p1*v1/10**3; \t#kJ/kg\n", "u2 = 2602; \t\t\t #kJ/kg\n", "\n", "# Results\n", "W = u1-u2; \n", "print (\"Work done = %.3f\")%(W),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done = 196.340 kJ/kg\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.58 page no : 201 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate per kg of steam :\n", "(i) The work done during expansion ;\n", "(ii) The heat flow to or from the cylinder walls during the expansion.\n", "'''\n", "\n", "# Variables\n", "p1 = 7.*10**5; \t\t\t#N/m**2\n", "x1 = 0.98;\n", "p2 = 0.34*10**5; \t\t#N/m**2\n", "vg = 0.273; \t\t\t#m**3/kg\n", "n = 1.1;\n", "v_g2 = 4.65; \t\t\t#m**3/kg\n", "u_f1 = 696.; \t\t\t#kJ/kg\n", "u_g1 = 2573.; \t\t\t#kJ/kg\n", "u_f2 = 302.; \t\t\t#kJ/kg\n", "u_g2 = 2472.; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "v1 = x1*vg;\n", "v2 = v1*(p1/p2)**(1/n);\n", "x2 = v2/v_g2;\n", "\n", "\n", "print (\"(i) Work done by the steam during the process\")\n", "W = (p1*v1-p2*v2)/(n-1)/10**3; \t\t\t#kJ/kg\n", "print (\"W = %.3f\")%(W), (\"kJ/kg\")\n", "\n", "\n", "print (\"(ii) Heat transferred\")\n", "u1 = (1-x1)*u_f1+x1*u_g1;\n", "u2 = (1-x2)*u_f2+x2*u_g2;\n", "Q = u2-u1 + W;\n", "print (\"Q = %.3f\")%(Q), (\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Work done by the steam during the process\n", "W = 450.232 kJ/kg\n", "(ii) Heat transferred\n", "Q = 169.289 kJ/kg\n" ] } ], "prompt_number": 55 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.59 page no : 203" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine the work done per kg of steam flow through the turbine\n", "'''\n", "\n", "# Variables\n", "p1 = 15.; \t\t\t#bar\n", "t1 = 350.; \t\t\t#0C\n", "C1 = 60.; \t\t\t#m/s\n", "p2 = 1.2; \t\t\t#bar\n", "C2 = 180.; \t\t\t#m/s\n", "s1 = 7.102; \t\t\t#kJ/kg\n", "s_f2 = 1.3609; \t\t\t#kJ/kg\n", "s_g2 = 7.2884; \t\t\t#kJ/kg\n", "h_f2 = 439.4; \t\t\t#kJ/kg\n", "h_fg2 = 2241.1; \t\t\t#kJ/kg\n", "h1 = 3147.5; \t\t\t#kJ/kg\n", "\n", "# Calculations\n", "x2 = (s1 - s_f2)/(s_g2-s_f2);\n", "h2 = h_f2+x2*h_fg2;\n", "W = (h1-h2) + (C1**2 - C2**2)/2/1000;\n", "\n", "# Results\n", "print (\"Work done = %.3f\")%(W),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done = 523.075 kJ/kg\n" ] } ], "prompt_number": 56 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.60 page no : 204" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine the quality of the steam leaving the nozzle.\n", "'''\n", "# Variables\n", "p1 = 10.; \t\t\t#bar\n", "t1 = 200.; \t\t\t#0C\n", "C1 = 60.; \t\t\t#m/s**2\n", "c2 = 650.; \t\t\t#m/s\n", "p2 = 1.5; \t\t\t#bar\n", "h1 = 2827.9; \t\t\t#kJ/kg\n", "h_f2 = 467.1; \t\t\t#kJ/kg\n", "h2 = 2618.45; \t\t\t#kJ/kg\n", "h_g2 = 2693.4; \t\t\t#kJ/kg\n", "\n", "# Calculations\n", "x2 = (h2-h_f2)/(h_g2-h_f2);\n", "\n", "# Results\n", "print (\"quality of steam leaving the nozzle = %.3f\")%(x2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "quality of steam leaving the nozzle = 0.966\n" ] } ], "prompt_number": 57 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.61 page no : 206" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the initial dryness fraction of the steam.\n", "'''\n", "\n", "# Variables\n", "h1 = 2776.4; \t\t\t#kJ/kg\n", "h2 = h1;\n", "h_f1 = 884.6; \t\t\t#kJ/kg\n", "h_fg1 = 1910.3; \t\t#kJ/kg\n", "\n", "# Calculations\n", "x1 = (h1-h_f1)/h_fg1;\n", "\n", "# Results\n", "print (\"Initial dryness fraction = %.3f\")%(x1)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Initial dryness fraction = 0.990\n" ] } ], "prompt_number": 58 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.62 page no : 207" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the exit condition of steam using Mollier chart.\n", "'''\n", "\n", "# Variables\n", "p1 = 10.; \t\t\t#bar\n", "x1 = 0.9; \t\t\t#bar\n", "p2 = 2.; \t\t\t#bar\n", "\n", "# Calculations\n", "# Umath.sing Mollier chart, we get\n", "x2 = 0.94;\n", "\n", "# Results\n", "print (\"x2 = \"),(x2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "x2 = 0.94\n" ] } ], "prompt_number": 59 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.63 Page no :208" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Change in entropy ;\n", "(ii) Change in enthalpy ;\n", "(iii) Change in internal energy.\n", "'''\n", "\n", "\n", "import math \n", "print (\"(a)From steam tables\")\n", "\n", "# Variables\n", "p1 = 15*10**5; \t\t\t#Pa\n", "p2 = 7.5*10**5; \t\t\t#Pa\n", "h_f1 = 844.7; \t\t\t#kJ/kg\n", "ts1 = 198.3; \t\t\t#0C\n", "s_f1 = 2.3145; \t\t\t#kJ/kg.K\n", "s_g1 = 6.4406; \t\t\t#kJ/kg.K\n", "v_g1 = 0.132; \t\t\t#m**3/kg\n", "h_fg1 = 1945.2; \t\t\t#kJ/kg\n", "x1 = 0.95;\n", "h_f2 = 709.3; \t\t\t#kJ/kg\n", "h_fg2 = 2055.55; \t\t\t#kJ/kg\n", "s_f2 = 2.0195; \t\t\t#kJ/kg\n", "s_g2 = 6.6816; \t\t\t#kJ/kg.K\n", "v_g2 = 0.255; \t\t\t#m**3/kg\n", "x2 = 0.9;\n", "x3 = 1;\n", "s_f3 = 0.521; \t\t\t#kJ/kg K\n", "s_g3 = 8.330; \t\t\t#kJ/kg K\n", "\n", "# Calculations\n", "h2 = h_f2+x2*h_fg2;\n", "h1 = h_f1 + x1*h_fg1;\n", "s1 = s_f1 + x1*(s_g1-s_f1);\n", "s2 = s1;\n", "ds_12 = s2-s1;\n", "\n", "s3 = s_f3+x3*(s_g3-s_f3);\n", "ds_23 = s3-s2;\n", "\n", "ds = 709.3 + 0.9 * 2055.55\n", "\n", "# Results\n", "print (\"(i) Change in entropy = %.3f\")% (ds), (\"kJ/kg K\")\n", "\n", "h3 = h2;\n", "\n", "dh = h2-h1;\n", "print (\"(ii) Change in enthalpy %.2f\")%(dh), (\"kJ/kg\")\n", "\n", "\n", "print (\"(iii) Change in internal energy\"),\n", "u1 = h1-p1*x1*v_g1/10**3;\n", "u2 = h2-p2*x2*v_g2/10**3;\n", "du = u2-u1;\n", "print (\"du = %.3f\")% (du), (\"kJ/kg\")\n", "\n", "\n", "\t\t\t# Only the expansion of steam from point 1 to 2 (i.e., isentropic expansion) is reversible because of unresisted flow whereas the expansion from point 2 to point 3 (i.e., throttling expansion) is irreversible because of frictional resismath.tance to flow. Increase of entropy also shows that expansion from point 2 to point 3 is irreversible.\n", "\n", "\n", "print (\"(b) Using Mollier chart\")\n", "h1 = 2692; \t\t\t#kJ/kg\n", "h2 = 2560; \t\t\t#kJ/kg\n", "s1 = 6.23; \t\t\t#kJ/kg K\n", "s2 = s1;\n", "s3 = 8.3; \t\t\t#kJ/kg K\n", "\n", "ds = s3-s1;\n", "print (\"(i) Change in entropy = %.3f\")%(ds), (\"kJ/kg K\")\n", "\n", "\n", "dh = h2-h1;\n", "print (\"(ii) Change in enthalpy = %.3f\")%(dh),(\"kJ/kg\")\n", "\n", "u3=u2-u1\n", "print (\"(iii) Change in internal energy =%.3f\")%(u3),(\"kJ/kg\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)From steam tables\n", "(i) Change in entropy = 2559.295 kJ/kg K\n", "(ii) Change in enthalpy -133.35 kJ/kg\n", "(iii) Change in internal energy du = -117.370 kJ/kg\n", "(b) Using Mollier chart\n", "(i) Change in entropy = 2.070 kJ/kg K\n", "(ii) Change in enthalpy = -132.000 kJ/kg\n", "(iii) Change in internal energy =-117.370 kJ/kg\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.64 Page no :212" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Calculate the mass of air which has left the receiver.\n", "\n", "\n", "# Variables\n", "V1 = 5.5; \t\t\t#m**3\n", "p1 = 16.*10**5; \t\t\t#Pa\n", "T1 = 315.; \t\t\t#K\n", "V2 = V1;\n", "p2 = 12.*10**5; \t\t\t#Pa\n", "R = 0.287*10**3;\n", "y = 1.4;\n", "\n", "# Calculations\n", "m1 = p1*V1/R/T1;\n", "T2 = T1*(p2/p1)**((y-1)/y);\n", "m2 = p2*V2/R/T2;\n", "\n", "# Results\n", "m = m1-m2;\n", "print (\"Mass of air which left the receiver = %.3f\")% (m), (\"kg\")\n", "\n", "# Note : Rounding error is there." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass of air which left the receiver = 18.081 kg\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.65 Page no :213" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine the work obtainable by utilising the kinetic energy of the discharge air to run a frictionless\n", "turbine\n", "'''\n", "# Variables\n", "cp = 1.; \t\t\t#kJ/kg.K\n", "cv = 0.711; \t\t\t#kJ/kg.K\n", "V1 = 1.6; \t\t\t#m**3\n", "V2 = V1;\n", "p1 = 5.*10**5; \t\t\t#Pa\n", "T1 = 373.; \t\t\t#K\n", "p2 = 1.*10**5; \t\t\t#Pa\n", "R = 287.;\n", "y = 1.4;\n", "\n", "# Calculations\n", "m1 = round(p1*V1/R/T1,2);\n", "T2 = round(T1*(p2/p1)**((y-1)/y),2);\n", "m2 = round(p2*V2/R/T2,3);\n", "KE = (m1*cv*T1)-(m2*cv*T2)-(m1-m2)*cp*T2;\n", "\n", "# Results\n", "print \"Kinetic energy of discharge air = %.3f\"% (KE), (\"kJ\")\n", "print (\"This is the exact answer when using proper value of cv\")\n", "\n", "# Book answer is wrong." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Kinetic energy of discharge air = 382.910 kJ\n", "This is the exact answer when using proper value of cv\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.66 Page no :214" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Final state condition ;\n", "(ii) Work done by the piston ;\n", "(iii) Heat transferred to oxygen.\n", "'''\n", "\n", "#For oxygen\n", "import math \n", "\n", "# Variables\n", "cpa = 0.88; \t\t\t#kJ/kg K\n", "Ra = 0.24; \t\t\t#kJ/kg K\n", "V1a = 0.035; \t\t\t#m**3\n", "p1a = 4.5; \t\t\t#bar\n", "T1a = 333.; \t\t\t#K\n", "V2a = 0.07; \t\t\t#m**3\n", "\n", "#For methane\n", "V1b = 0.07; \t\t\t#m**3\n", "V2b = 0.035; \t\t\t#m**3\n", "p1b = 4.5; \t\t\t#bar\n", "T1b = 261; \t\t\t#K\n", "cpb = 1.92; \t\t\t#kJ/kg K\n", "Rb = 0.496; \t\t\t#kJ/kg K\n", "\n", "# Calculations and Results\n", "yb = cpb/(cpb-Rb); \t\t\t#for methane\n", "cva = cpa-Ra; \t\t\t#for oxygen\n", "\n", "print (\"(i) Final state condition\")\n", "\n", "p2b = p1b*(V1b/V2b)**yb;\n", "print (\"p2 for methane = %.3f\")% (p2b), (\"bar\")\n", "\n", "T2b = p2b*V2b*T1b/p1b/V1b;\n", "print (\"T2 for methane = %.3f\")% (T2b), (\"K\")\n", "\n", "p2a = p2b;\n", "\n", "T2a = p2a*V2a/p1a/V1a*T1a;\n", "print (\"T2 for oxygen = %.3f\")% (T2a), (\"K\")\n", "\n", "Wb = (p1b*V1b - p2b*V2b)/(yb-1)*100; \t\t\t#kJ\n", "\n", "print (\"(ii)The piston will be in virtual equilibrium and hence zero work is effected by the piston.\")\n", "\n", "Wa = -Wb;\n", "\n", "ma = p1a*V1a/Ra/T1a*10**2;\n", "\n", "Q = ma*cva*(T2a-T1a) + Wa;\n", "print \"(iii) Heat transferred to oxygen = %.3f\"% (Q), (\"kJ\")\n", "\n", "# Rouding error is there." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Final state condition\n", "p2 for methane = 11.458 bar\n", "T2 for methane = 332.272 K\n", "T2 for oxygen = 1695.733 K\n", "(ii)The piston will be in virtual equilibrium and hence zero work is effected by the piston.\n", "(iii) Heat transferred to oxygen = 196.572 kJ\n" ] } ], "prompt_number": 4 } ], "metadata": {} } ] }